Compact support gives an entire Fourier-Laplace transform by slices

Fix a representative. If $K=\mathrm{\varnothing }$ then $f=0$ almost everywhere, so every integral below vanishes by [F3], giving $F\equiv 0$, and all claims of the lemma hold trivially; assume $K\mathrm{\ne }\mathrm{\varnothing }$ and let $M$ be the bound of [F5]. Multiplying $f$ by the indicator of the closed set $K$ produces a measurable representative in $L^1$ that vanishes everywhere outside $K$ and agrees with $f$ almost everywhere, so no integral below changes ([F3, F6]); from here we use this representative.

Absolute convergence and representative independence. For $x\in K$ the exponential law and modulus formula [F1] give $\mathrm{\mid }f(x)e^{-2\pi ix\cdot z}\mathrm{\mid }=\mathrm{\mid }f(x)\mathrm{\mid }{e}^{2\pi x\cdot \mathrm{Im}z}\le \mathrm{\mid }f(x)\mathrm{\mid }{e}^{2\pi M\mathrm{\mid }\mathrm{Im}z\mathrm{\mid }}$, while for $x\mathrm{\notin }K$ both sides vanish. Hence ${\int }_{{\mathbb{R}}^n}\mathrm{\mid }f(x)e^{-2\pi ix\cdot z}\mathrm{\mid }dx\le e^{2\pi M\mathrm{\mid }\mathrm{Im}z\mathrm{\mid }}\mathrm{\parallel }f{\mathrm{\parallel }}_1<\mathrm{\infty }$ for the given $z$, and since $z$ was arbitrary the integral defining $F$ converges absolutely at every point of ${\mathbb{C}}^n$. If $\tilde{f}=f$ almost everywhere is a second representative, the two integrands agree almost everywhere and are both integrable, so the two integrals agree by [F3]; thus the value is representative-independent.

The slice as a one-variable integral. Fix $j$ and complex numbers $z_k$ for $k\mathrm{\ne }j$, and set $g(x):=f(x)e^{-2\pi i{\sum }_{k\mathrm{\ne }j}{x}_k{z}_k}$. Then $g$ is measurable and $\mathrm{\mid }g(x)\mathrm{\mid }\le \mathrm{\mid }f(x)\mathrm{\mid }{e}^{2\pi M\mathrm{\mid }\mathrm{Im}z\mathrm{\mid }}$ for every $x$ by the same bound as in step 1.2, so $g\in L^1$; for $w\in \mathbb{C}$ define $h(w):={\int }_{{\mathbb{R}}^n}g(x)e^{-2\pi i{x}_{j}w}dx$. By the addition law [F1] the integrand equals $f(x)e^{-2\pi ix\cdot z^{\mathrm{\prime }}}$ with $z^{\mathrm{\prime }}=(z_1,\dots ,z_{j-1},w,z_{j+1},\dots ,z_n)$, so $h(w)=F(z^{\mathrm{\prime }})$ and step 1.2 shows that the integral for $h(w)$ converges absolutely at every $w\in \mathbb{C}$.

Difference quotients are integrals. Fix $w\in \mathbb{C}$ and $s\in \mathbb{C}\setminus \{0\}$. For every $x$ the addition law [F1] gives $e^{-2\pi i{x}_j(w+s)}-e^{-2\pi i{x}_{j}w}=e^{-2\pi i{x}_{j}w}(e^{-2\pi i{x}_{j}s}-1)$, and both integrands $g(x)e^{-2\pi i{x}_j(w+s)}$ and $g(x)e^{-2\pi i{x}_{j}w}$ are integrable because $g\in L^1$ and the exponential factors are bounded on the support of $g$ by step 2.1. Applying additivity and scaling from [F3] therefore yields $\frac{h(w+s)-h(w)}{s}={\int }_{{\mathbb{R}}^n}g(x)e^{-2\pi i{x}_{j}w}\frac{e^{-2\pi i{x}_{j}s}-1}{s}dx\mathrm{.}$

Uniform remainder estimate. Put $S:=S_j\le M$ and $L:=\int g(x)(-2\pi i{x}_j)e^{-2\pi i{x}_{j}w}dx$, which is absolutely convergent since $\mathrm{\mid }{x}_j\mathrm{\mid }\le S$ on the support of $g$. If $S=0$, step 3.1 is identically zero and $L=0$. Otherwise, for any $\epsilon >0$, [F1] gives $\delta >0$ such that $\mathrm{\mid }(e^{\zeta }-1)\mathrm{/}\zeta -1\mathrm{\mid }<\epsilon$ when $0<\mathrm{\mid }\zeta \mathrm{\mid }<\delta$. For $0<\mathrm{\mid }s\mathrm{\mid }<\delta \mathrm{/}(2\pi S)$, setting $\zeta =-2\pi i{x}_{j}s$ therefore bounds the difference between the quotient integrand and its limiting integrand by $2\pi S\epsilon e^{2\pi S\mathrm{\mid }\mathrm{Im}w\mathrm{\mid }}\mathrm{\mid }g(x)\mathrm{\mid }$ on $K$, with zero remainder when $x_j=0$. By [F2] and linearity [F3], $\mid \frac{h(w+s)-h(w)}{s}-L\mid \le 2\pi S\epsilon e^{2\pi S\mathrm{\mid }\mathrm{Im}w\mathrm{\mid }}\mathrm{\parallel }g{\mathrm{\parallel }}_1\mathrm{.}$ Since $\epsilon$ is arbitrary, the complex difference quotient tends to $L$ directly, without selecting a sequence.

Conclusion. The limit in step 4.1 shows that $h$ is complex differentiable at every $w\in \mathbb{C}$ with derivative $h^{\mathrm{\prime }}(w)={\int }_{{\mathbb{R}}^n}g(x)(-2\pi i{x}_j)e^{-2\pi i{x}_{j}w}dx$; a function holomorphic on all of $\mathbb{C}$ is entire by definition ([F7]). Rewriting the derivative integrand gives the displayed formula for $\mathrm{\partial }F\mathrm{/}\mathrm{\partial }{z}_j$. At real $z=\xi \in {\mathbb{R}}^n$ the defining integral of $F$ is literally the $L^1$ transform formula, so $F(\xi )=\widehat{f}(\xi )$ at every real $\xi$ by [F4].